A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the dist

Question

A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance x 2 from the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity 3 m s - 1 . The total initial energy of the spring is:

TestHub · Accepted Answer

By mechanical energy conservation between compression positions x and x 2 1 2 k x 2 = 1 2 k x 2 2 + 1 2 m v 2 1 2 k x 2 - 1 2 k x 2 4 = 1 2 m v 2 1 2 k x 2 3 4 = 1 2 m v 2 v = 3 k x 2 4 m = 3 k m x 2 On collision with a block at rest ∵ Velocities are exchanged ⇒ elastic collision between identical masses. ∴ v = 3 = 3 k m x 2 ⇒ 6 = 3 k m x ⇒ x = 6 m 3 k ∴ The initial energy of the spring is U = 1 2 k x 2 = 1 2 k × 36 m 3 k = 6 m U = 6 × 0.1 = 0 .6 J

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